Let's find the slope [tex]\( (m) \)[/tex] and the y-intercept [tex]\( (b) \)[/tex] of the trendline for the given experimental data.
### Step-by-Step Solution:
1. Given Data:
- Pressures (atm): 1.05, 1.10, 1.15, 1.35, 1.45
- Masses (grams): 0.127, 0.133, 0.139, 0.151, 0.163
- Volumes (mL): 9.10, 18.20, 27.30, 45.50, 63.70
- Temperature: 21.5°C
2. Constants and Conversions:
- Ideal gas constant [tex]\( R \)[/tex] = 0.0821 L·atm/(K·mol)
- Molar mass of methane (CH₄) ≈ 16.04 g/mol
3. Convert Mass to Moles:
The number of moles ([tex]\( n \)[/tex]) of methane is calculated using the formula:
[tex]\[
n = \frac{\text{mass}}{\text{molar mass}}
\][/tex]
Applying this to each mass in the data:
[tex]\[
\begin{aligned}
n_1 & = \frac{0.127}{16.04} \approx 0.00792 \\
n_2 & = \frac{0.133}{16.04} \approx 0.00829 \\
n_3 & = \frac{0.139}{16.04} \approx 0.00867 \\
n_4 & = \frac{0.151}{16.04} \approx 0.00941 \\
n_5 & = \frac{0.163}{16.04} \approx 0.01016 \\
\end{aligned}
\][/tex]
4. Plot the Data:
On the x-axis, plot the number of moles ([tex]\( n \)[/tex]) and on the y-axis, plot the corresponding volumes (mL):
[tex]\[
\begin{aligned}
& (0.00792, 9.10) \\
& (0.00829, 18.20) \\
& (0.00867, 27.30) \\
& (0.00941, 45.50) \\
& (0.01016, 63.70) \\
\end{aligned}
\][/tex]
5. Linear Regression:
Perform a linear regression to find the slope ([tex]\( m \)[/tex]) and y-intercept ([tex]\( b \)[/tex]) of the line that best fits the plotted data.
6. Result:
After performing the regression analysis:
- Slope [tex]\( (m) \)[/tex]: [tex]\( 24327.33 \, \text{mL/mol} \)[/tex]
- Y-intercept [tex]\( (b) \)[/tex]: [tex]\( -183.52 \, \text{mL} \)[/tex]
So, the slope [tex]\( (m) \)[/tex] is approximately [tex]\( 24327.33 \, \text{mL/mol} \)[/tex] and the y-intercept [tex]\( (b) \)[/tex] is approximately [tex]\( -183.52 \, \text{mL} \)[/tex].
